1.2 CHAPTER 1 : Relation and Functions [TOPIC 1] …...1.2 CHAPTER 1 : Relation and Functions [TOPIC 1] Concept of Relations and Function Summary Relation Definition: If (a, b) R,

CHAPTER 1 : Relation and Functions1.2

[TOPIC 1] Concept of Relations and Function

Summary

RelationDefinition: I f (a, b) R, we say that a is related to bunder the relat ion R and we wr ite i t as aRb.

Domain of a relat ion: The set of fi rst components ofal l the ordered pairs which belong to R is the domainof R.

Domain : ,R a A a b R b B

Range of a relat ion: The set of second components ofal l the ordered pairs which belong to R is the domainof R.

Range of : ,R b B a b R a A

Types of relations: Empty relation: Empty relat ion is the relat ion R

from X to Y i f no element of X is related to anyelement of Y, i t is given by .R X Y

For example, let 2,4,6 , 8,10,12X Y

, : , and is oddR a b a X b Y a b

R is an empty relat ion.

Universal relation:Universal relat ion is a relat ionR from X to Y i f each element of X is related toevery element of Y i t is given by R = X Y.

For example, let X = {x, y}, Y = {x, z}

R = {(x, x), (y, z), (y, x), (y, z)}

R = X Y, so relat ion R is a universal relat ion.

Reflexive relation: Reflexive relat ion R in X is arelat ion with (a, a) R a X.

For example, let X = {x, y, z} and relat ion R is givenas

R = {(x, x), (y, y), (z, z)}

Here, R is a reflexive relat ion on X.

Symmetric relation: Symmetr ic relat ion R in Xis a relat ion sat isfying (a, b) R implies (b, a) R.

For example, let X = {x, y z} and relat ion R is given as

R = {(x, y), (y, x)}

Here, R is a symmetr ic relat ion on X. Transitive relation: Transit ive relat ion R in X is

a relat ion sat isfying (a, b) R and (b, c) R impliesthat (a, b) R.


R = {(x, z), (z, y), (x, y)}

Here, R is a t ransit ive relat ion on X. Equivalence relation: I t is a relat ion R in X which

is reflexive, symmetr ic and t ransit ive.


R = {(x, y), (x, x), (y, x), (y, y), (z, z), (x, z), (z, x), (y, z)}

Here, R is reflexive, symmetr ic and t ransit ive. SoR is an equivalence relat ion on X.

Equivalence class [a] containing a X for anequivalence relat ion R in X is t he subset of Xcontaining al l elements b related to a.

FunctionDefinition: A rule f which associates each element ofa non-empty set A with a unique element of anothernon-empty set B is cal led a funct ion.

Types of functions: I njective function: A funct ion f : X Y is one-

one (or inject ive) i f f(x1) = f (x2) x1 = x2x1, x2 X.

Surjective function: A funct ion f : X Y is onto (orsur ject ive) i f given any y Y, x X such thatf(x) = y.

Bijective function: A funct ion f : X Y is one-oneand onto (or biject ive), if f is both one-one and onto.

Composite function: The composit ion of funct ionsf : A B and g : B C is the funct ion gof : A Cgiven by gof(x) = g(f(x)) x A

Invertible function: A function f : X Y is invertiblei f g : Y X such t hat gof = I X andfog = I Y.

A funct ion f : X Y is inver t ible i f and only i f f isone-one and onto.

Steps to find inverse of a function

Let f (x) = y where x X and y Y

Solve f (x) = y for x in terms of y.Now replace x with f – 1(y) in the expression obtainfrom the above step.

Finally to find the inverse funct ion of f f– 1(x) replacey with x in the expression obtained from the abovestep.

s

CHAPTER 1 : Relation and Functions1.12

Summary Binary Operation: A binary operat ion * on a set A is a funct ion * from A A to A. We denote * (a, b) by a * b.

An element e X is the identity element for binary operat ion * : X X X, i f a * e = a = e * aa X.

An element a X is invert ible for binary operat ion * : X X X, i f there exists b X such thata * b = e = b * a where, e is the ident i ty for the binary operat ion * . The element b is cal led inverse of a andis denoted by a– 1.

An operat ion * on X is commutative i f a * b = b * aa, b in X.

An operat ion * on X is associative i f

(a * b) * c = a * (b * c)a, b, c in X.

[TOPIC 2] Binary Operations

PREVIOUS YEARS’EXAMINATION QUESTIONSTOPIC 2

1 Mark Questions

1. Let * be a binary operat ion, on the set of all non

zero real numbers, given by *5ab

a b for al l

, 0a b R . Find the value of x, given that

2* * 5 10x .

[DELH I 2014]

2. Let * be a ‘binary operat ion on N given a * b =

LCM (a, b) for all a,b N . find 5*7 .

[ALL I NDI A 2012]

3. L et * be a bi nar y oper at i on on N gi ven bya * b = LCM (a, b) for al l a,b N. Find 5*7.

[DELH I 2012]

4. I f a* b denote the larger of ‘a' and ‘b' and i f

3a b a b , t hen wr i t e t he val ue of

5 10 , where * and O are binary operat ions.

[DELH I 2018]

5. Consi der t he bi nar y oper at i on * on t he set{1, 2, 3, 4, 5} defined by a*b = min {a, b} wr i te theoperat ion table of the operat ion .

[DELH I 2011]

6 Marks Questions6. Let A = Q × Q, where Q is the set of al l rat ional

numbers, and be a binary operat ion defined on Aby (a, b) * (c, d) = (ac, b + ad), for all (a, b) (c, d) A.

Find

(i) the ident i ty element in A.

(i i ) the inver t ible element of A.

[ALL I NDI A 2015]

7. Discuss the Commutat ivi ty and associat ivi ty ofbinary operat ion * defined on A = Q – {1} by the

r ule ,a b a b ab a b A . Also f i nd t he

ident i ty element of * in A and hence find theinver t ible elements of A.

[DELH I 2017]

Solutions1. Given that , 2 * x * 5 10

52 * 10

5x

*

5ab

a b [½]

2 * 10x

210

5x

510 25

2x [½]

2. Given a*b = LCM (a, b)

5 * 7 = LCM (5, 7) = 35 [1]

CHAPTER 2 : Inverse Trigonometric Functions2.16

Summary

Definition of inverse trigonometricfunctions:I nverse t r igonomet r ic funct ions are the inverse oft r igonometr ic funct ions, we can represents them byusing “ -1” or arc on t r igonometr ic funct ions. Also theRange of t r igonometr ic funct ion becomes the Domainof I nverse t r igonometr ic funct ion.

For ex: sinx y wi l l be represented as arcsiny x or1siny x .

Range of x = sin y is [– 1, 1] and Domain of y = arcsin xis [– 1, 1].

The inverse t r igonometr ic funct ions are also cal ledas I nverse Circular Functions.

Function: y = sin – 1xDomain: [– 1, 1]

Range:

,

2 2

Y

52

32

2

2 1

2–

X XO–1

– 32–

52–

– 2

Yy x = sin –1

Function: y = cos– 1xDomain: [– 1, 1]

Range: [0, ]

Inverse Trigonometric Functions

Y

52

32

2

2

1

2–

X XO–1

–

32–

52–

– 2

Yy x = cos –1

Function: y = cosec– 1xDomain: R – (– 1, 1)

Range: , 0

2 2

Y

32

2

2

12–

X XO

–1

–

Yy x = cosec –1

–2

2

CHAPTER 2 : Inverse Trigonometric Functions 2.17

Function: y = sec– 1xDomain: R – (– 1, 1)

Range:

0,2

Y

32

2

2

1

2–

XO

–

Yy x = sec –1

X2–2 –1

Function: y = tan – 1xDomain: R

Range:

,

2 2

Y

232

2

O–1–2

X X1 2

2–

–

Yy x = tan –1

Function: y = cot – 1x

Domain: R

Range: (0, )

Y

32

2

2

12–

X O–1

–

Yy x = cot –1

–2

2X

Properties:

1sin sin x x

1cos cos x x

1tan tan x x

1cosec cosec x x

1sec secx x

1cot cot x x

1 1 1

sin cosec , 1,1x xx

1 1 1

cosec sin , , 1 1,x xx

1 1 1

cos sec , 1,1x xx

1 1 1

sec cos , , 1 1,x xx

1

1

1

1cot , 0

tan1

cot , 0

xx

xx

x

1

1

1

1tan , 0

cot1

tan , 0

xx

xx

x

CHAPTER 2 : Inverse Trigonometric Functions2.18

1 1sin cos , 1,12

x x x

1 1tan cot , R2

x x x

1 1cosec sec , 12

x x x

1 1sin sinx x

1 1cos cosx x

1 1tan tanx x

1 1cot cotx x

1 1sec secx x

1 1cosec cosecx x

1 1 1tan tan tan

1x y

x yxy

1 1 1tan tan tan1x y

x yxy

12

21 1

2

12

2sin , 1

1

12tan cos , 0

1

2tan , 1 1

1

xx

x

xx x

x

xx

x


1 Mark Questions

1. Write the pr incipal value of

1 1 1

tan 1 cos .2

[ALL I NDI A 2013]

2. Write the value of

1 1

tan 2tan .5

[DELH I 2013]

3. Wr i t e t he pr i nci pal val ue of

1 11 1

cos 2sin .2 2

[DELH I 2012]

4. Write the value of 1 1

sin sin3 2

[DELH I 2011]

5. Write the pr incipal value of 1t an 1

[ALL I NDI A 2011]

6. Write the pr incipal value of

1 1tan 3 cot 3 .

[ALL I NDI A 2013, 2018]

7. Write the value of 1 1 3

tan 2sin 2cos .2

[All I ndia 2013]

8. I f 1 1tan tan , 1,4

x y xy t hen wr i t e t he

value of x + y + xy.

[ALL I NDI A 2014]

4 Marks Questions9. Find the value of the fol lowing:

21 1

2 21 2 1

tan sin cos , 1,y 02 1 1

x yx

x y

and 1.xy

[DELH I 2013]

10. I f 1 1sin cot 1 cos tan ,x x then find x.

[DELH I 2015]

11. Prove that

1 11 1 2tan cos tan cos .

4 2 4 2

a a bb b a

[DELH I 2017]

CHAPTER 3 : Matrices3.32

[TOPIC 1] Matrix and Operations on Matrices

Summary A matr ix i s an or der ed r ect angular ar r ay of

numbers or funct ions. The numbers are cal led theelements of the matr ix.

11 12 13 1

21 22 23 2

31 32 33 3

1 2 3

.... ....

... ...

... ...

. . . ... .... .

. . . ... ... .

.... ...

n

n

n

m m m mn m n

a a a aa a a aa a a a

A

a a a a

Example:

14 2

20 3 5

1 6 7

A

The order of the matr ix is determined by m nwhere m is the number of rows and n is the numberof columns.

The matr ix with m n order can be represented

as

; ,i j m nA a i j also 1 , 1 ,i m j n .

Types of Matrices

Column matrix i s a mat r ix which has only 1column. I t is defined as A = [ai j]m1.

Example:

3 1

2

3

4

Row matrix is a matr ix which has only row. I t isdefined as B = [bi j]1n.

Example:

1 32 1 3

The square matrix is the matr ix which has equalnumber of rows and columns i .e. the mat r ix inwhich m = n. I t is defined as A = [ai j]mm.

Example:

3 3

3 9 1

7 6 3

9 0 2

The squar e mat r i x A = [ai j ]mm i s said t o bea diagonal matr ix i f al l i ts non-diagonal elementsare zero. I t is defined as A = [ai j]mm i f ai j = 0, wheni j.

A scalar matrix is the one in which the diagonalelements of a diagonal matrix are equal. I t is definedas A = [ai j]mm i f ai j = 0, when i j and ai j = k , wheni = j, where k is some constant .

Example:

3 0

0 3

I dentity matrix is the square matr ix where thediagonal elements are al l 1 and rest are al l zero. I tis defined as A = [ai j]mm where ai j = 1 i f i = j andai j = 0 i f i j.

Example:

2 2

1 0

0 1

A zero matrix is the one in which al l the elementsare zero.

Example:

3 3

0 0 0

0 0 0

0 0 0

Two matr ices A = [ai j] and B = [bi j] are equal i f theyare of the same order and also each element ofmatr ix A is equal to the cor responding element ofMatr ix B.

The sum of the two mat r ices A = [ai j ] andB = [b i j ] of same or der m n i s def i ned asC = [ci j]mnwhere ci j = ai j + bi j.

I f x is a scalar and A = [ai j]mn is a matr ix, then xA isthe matr ix obtained by mult iplying each elementof the matr ix by the scalar x. I t can be defined asxA = x[ai j]mm = [x(ai j)]mn.

– A denotes the negat ive of a matr ix. – A = (– 1)A.

The difference of the two matrices A = [ai j] andB = [bi j]of same order m n is defined as C = [ci j]mnwhere ci j = ai j – bi j.

I f A = [ai j] and B = [bi j] are matr ices of the sameorder, say m n then A + B = B + A. I t is cal led thecommutative law.

CHAPTER 3 : Matrices 3.33

I f A = [ai j], B = [bi j] and C = [ci j] are the threematr ices of same order, say m n, then (A + B) + C= A + (B + C). I t is cal led the associative law.

I f B = [bi j] is a matr ix of order m n and O is a zeromatr ix of the order m n, then B + O = O + B = B.O is the addit ive ident i ty for matr ix addit ion.

I f B = [bi j] is a matr ix of order m n then we haveanother matr ix as – B = [– bi j] of the order m nsuch that B + (– B) = (– B) + B = O. So, – B is theaddit ive inverse of B.

I f A = [ai j] and B = [bi j] are matr ices of the sameorder, say m n, and x and y are the scalars, then

x(A + B) = xA + xB

(x + y)A = xA + yA

I f A = [ai j] is a matr ix of order m n and B = [bjk]is a mat r ix of order n p t hen the product ofthe matr ices A and B is a matr ix C of order m p.

I t can be denoted as AB = C = [Cik]mp, where

1

n

ik i j jkj

c a b

Propert ies of mult iplicat ion of matr ices areas fol lows:

The associat ive law: I f there are 3 matr icesX, Y and Z we have (XY)Z = X(YZ)

Dist r ibut ive law: I f there are 3 matr ices X, Yand Z then:

X(Y + Z) = XY + XZ

(X + Y)Z = XZ + YZ

The exist ence of mul t i pl i cat i ve ident i t y:For every square matr ix X, there exists anident i ty matr ix of the same order such thatIX = XI = X.

Mult ipl icat ion is not commutat ive: AB BA


1 Mark Questions

1. Find the value of a i f

2 1 5.

2 3 0 13a b a ca b c d

[DELH I 2013]

2. I f

1 1 4 1,

3 2 1 3x xx x then wr ite the value

of x.

[DELH I 2013]

3. I f

9 1 4 1 2 1,

2 1 3 0 4 9A t hen f i nd

the matr ix A.

[DELH I 2013]

4. Write the element 23a of a 3 3 matr ix i jA a

whose elements i ja are given by

2i j

i ja .

[DELH I 2015]

5. I f A is a 3 3 inver t ible matr ix, then what wi l l

be the value of k i f

1det det kA A .

[DELH I 2017]

6. I f A is a square matr ix such that 2A =I , then find

the simpli fied value of

3 3A-I + A+I – 7A

[DELH I 2016]

7. I f

2 3 1 3 4 6,

5 7 2 4 9 x wr i te the value

of x.

[ALL I NDI A 2012]

8. For a 2 2 matr ix, A = i ja , whose elements

are given by ,i j

ia

j wr ite the value of 12a

[DELH I 2011]

9. For what value of x the matr ix

5 12 4

x x is

singular?

[DELH I 2011]


Summary The t ranspose of the mat r ix A = [a i j]mn i s

denot ed by AT = [a j i ] nm and i s obt ai ned byinterchanging the rows with columns of matr ix A.

Example: I f

4 1

2 0A , then

4 2

1 0TA .

Some proper t ies of t ranspose of the matr ices areas fol lows:

I f A and B are matr ices of suitable orders then

(AT)T = A

(kA)T = kAT (Where k is any constant)

(A + B)T = AT + BT

(AB)T = BTAT

I f AT = A then the square matr ix A = [ai j] is said to besymmetric matrix for all possible values of i and j.

Example:

1 2 4

2 3 7

4 7 0

[TOPIC 2] Transpose of a Matrix and Symmetric andSkew Symmetric Matrices

I f AT = – A then the square matr ix A = [ai j] is said tobe skew symmetric matrix for al l the possiblevalues of i and j . All the diagonal elements of askew symmetr ic matr ix are zero.

Example:

1 2 4

2 3 7

4 7 0

A then

1 2 4

2 3 7

4 7 0

TA

1 2 4

2 3 7

4 7 0

A

For any square matr ix A with real number entr ies,A + AT is a symmetr ic matr ix and A – AT is a skewsymmetr ic matr ix.

Any square matr ix can be expressed as the sum ofa symmetr ic matr ix and a skew symmetr ic matr ix

i .e 1 12 2

T TA A A A A


1 Mark Questions

1. M at r i x

0 2 2

3 1 3

3 3 1

bA

a i s gi ven t o be

symmetr ic, find values of a and b.

[DELH I 2016]

2. I f the matr ix

2 1

1

0 30

0

Aa

b a skew symmetr ic

matr ix, find the value of ‘a' and ‘b'?

[DELH I 2018]

3. For what val ue of x, i s t he mat r i x

0 1 2

1 0 3

3 0

Ax

a symmetr ic matr ix?

[ALL I NDI A 2013]

4. I f A is a skew-symmetr ic matr ix of order 3, thenprove that det A = 0.

[ALL I NDI A 2017]

5. I f A =

cos sin

sin cos

, find 0

2

sat isfying

22TA A I when is t ranspose of TA A .

[ALL I NDI A 2016]

2 Marks Question6. Show that al l the diagonal elements of a skew

symmetr ic matr ix are zero.

[DELH I 2017]

e


Elementary operat ion of a matr ix are as fol lows:

I nterchanging of two rows or columns:

Ri Rj or Ci Cj represents that the i th row

or column is inter changed wi th j t h r ow or

column.

M ultiplying the row or column of matrix

by non- zero scalar: Ri lRj or Ci lCj

where I is any non- zero number, represents

the i th row or column is mult ipl ied by I .

Adding t he element s of any r ow or

column to another row or column: Ri

Ri + lRj or Ci Ci + lCj, where I is any non-

zero number, represents that the j th row or

[TOPIC 3] Inverse of matrices by Elementary rowtransformation

col umn i s mul t i pl i ed by I and added t o

respect ive element of i th row or column.

I f X i s a squar e mat r i x of or der n and i f

t her e exi st s anot her squar e mat r i x Y of t he

same order n, such that XY = YX = I , then Y is

cal led the inverse matr ix of X and i t is denoted by

X– 1.

The inverse of a matr ix can be found using row or

column operat ions.

I f Y is the inverse of X, then X is also the inverse of

Y. Also, (XY)– 1 = Y– 1X– 1

I nverse of a square matr ix, i f i t exists, is unique.


1 Mark Question

1. I f 5 3 82 0 1 ,

1 2 3 wr i t e t he m i nor of t he

element a23.

[DELH I 2012]

6 Marks Questions2. Use elementary transformat ions, find the inverse

of the matr ix

8 4 32 1 1

1 2 2

A

And use i t to solve the fol lowing system of l inearequat ions:

8 4 3 19 2 5 2 2 7x y z x y z x y z

[DELH I 2016]

3. Using element ar y t r ansfor mat ions, f i nd t he

inverse of the matr ix

1 3 2

3 0 1

2 1 0

[DELH I 2011]

4. Using elementary operat ions, find the inverse ofthe fol lowing matr ix:

1 1 2

1 2 3

3 1 1

[DELH I 2012]

5. Using elementary row t ransformat ions, find the

inverse of the matr ix

1 2 3

2 5 7

2 4 5

A .

[DELH I 2018]

6. Find A-1 using row elementary operat ions, given

that

2 0 1

5 1 0

0 1 3

A .

[DELH I 2011]

CHAPTER 4 : Determinants4.48

Summary Definition: A determinant is a number (real or

complex) that can be related to any square matr ixA = [ai j] of order n. I t is denoted as det(A).

Determinant of a mat r ix 11 12

21 22

a aA

a a

can be

given as:

11 1211 22 12 21

21 22

a aA a a a a

a a

Determinant of a matr ix 11 12 13

21 22 23

31 32 33

a a aA a a a

a a a

by

expanding along R1 can be given as:

[TOPIC 1] Expansion of Determinant

22 231 111

32 33

det ( 1)a a

A A aa a

21 231 212

31 33

( 1)a a

aa a

21 221 3

1331 32

( 1)a a

aa a

Minors: The minor M i j of ai j in A is the determinantof the square sub matr ix of order (n – 1) obtainedby delet ing i ts i th row and j th column in which ai jl ies. I t is denoted by M i j.

M inor of an element of a determinant of ordern(for al l n 2) is a determinant of order n – 1.

Co-factors: Co-factor of an element ai j is definedby A i j = (– 1)(i + j)M i j, where M i j is a minor of ai j.Co-factor is denoted by A i j.


1 Mark Questions

1. I f 3x 7 8 7

,2 4 6 4

Find the value of x.

[ALL I NDI A 2014]

2. I f i jA i s t he cofact or of el emen t i ja of t he

determinant

2 3 5

6 0 4 ,

1 5 7then wr ite the value of

32 32. .a A

[ALL I NDI A 2013]

3. x x 3 4

,1 x 1 2

wr ite the posit ive value of x

[ALL I NDI A 2011]

4. I f 5 3 8

2 0 1 ,

1 2 3 wr i te the minor of the elements

a23.

[ALL I NDI A 2012]

5. Simpli fy:

cos sin sin coscos sin

sin cos cos sin

[DELH I 2012]

6. I f

2 5 6 2

,8 7 3

xx wr i te the value of x.

[DELH I 2014]

7. Find the maximum value of

1 1 11 1 sin 1

1 1 1 cos

[DELH I 2016]

s

CHAPTER 4 : Determinants4.50

[TOPIC 2] Properties of Determinants

Summary Propert ies of determinants:

The val ue of t he det er m i nan t r emai nsunchanged i f i t s r ows and col umns ar einterchanged.

I f A is a square matr ix, then det (A) = det (A),where A= t ranspose of A.

I f any two columns (or rows) of a determinantar e i n t er changed, t hen t he si gn of t hedeterminant changes.

The determinant of the product of two squarematr ices of same order is equal to the productof t hei r r espect i ve det er mi nant s, t hat i sAB = AB.

I f any two rows or columns of a determinantare ident ical (i .e. al l cor responding elementsare same), then value of determinant is zero.

I f each element of a r ow or a column of adeterminant is mult ipl ied by a constant k, thenits value gets mult ipl ied by k .

I f some or al l elements of a row or column of adeterminant are expressed as sum of two ormor e t er ms, t hen t he det er minant can beexpressed as sum of two or more determinants.

For example:

1 2 3

1 2 3

1 2 3

a x a y a zb b bc c c

1 2 3

1 2 3 1 2 3

1 2 3 1 2 3

a a a x y zb b b b b bc c c c c c

I f we mul t iply each element of a r ow or acolumn of a determinant , by a constant k, thenthe value of the determinant is also mult ipl iedby k.

Mult iplying a determinant by k means mult iplyelements of any one row or any one columnby k.

Addi ng or subt r act i ng each el emen t ofanycolumn or any row of a determinant withtheequimult iples of corresponding elements of anyother r ow or column, does not changes thevalue of the determinant , i .e., t he value ofdet erminant r emain same i f we apply t heoperat ion Ri Ri kRj.

Area of triangle: I f a t r iangle is given with i tsver t ices at points (x1, y1), (x2, y2) and (x3, y3), thenits area can be calculated as:

1 1

2 2

3 3

11

12

1

x yx yx y

Area is a posit ive quant i ty, so we always takethe absolute value of the determinant .

I f the area of the t r iangle is already given, usebot h posi t i ve and negat i ve val ues of t hedeterminant for calculat ion.

Area of t r iangle formed by three coll inear pointsis always zero.


1 Mark Question1. I f A is a 3 3 matr ix and| 3A| = k| A| , then wr ite

the value of k.

[ALL I NDI A 2016]

4 Marks Questions2. Using the proper t ies of determinants, prove the

fol lowing :

1 12 ( 1) ( 1)

3 (1 ) 1 ( 2) 1 ( 1)

x xx x x x x

x x x x x x x x

2 26 (1 )x x

[ALL I NDI A 2015]

CHAPTER 4 : Determinants 4.57

[TOPIC 3] Adjoint and Inverse of a Matrix

Summary Adjoint of a square matrix:

The adjoint of a square matr ix A = [ai j]nn is definedas the t ranspose of the matr ix [A i j]nn, where A i j isthe cofactor of the element ai j. Adjoint of the matr ixA is denoted by "adj(A)".

Example:

1 4

2 3A then,

3 4

2 1adj A

I f A be any given square matr ix of order n, thenA(adj A) = (adj A)A = AI , where I is theident i ty matr ix of order n.

I nverse of a M atrix

Singular matr ix: I t i s a mat r i x wi th zerodeterminant value. i .e. A = 0.

Non-singular matrix: I t is a mat r ix wi th anon-zero determinant value. i .e. A 0.

I f A andB are nonsingular matr ices of the sameorder, then AB and BA are also nonsingularmatr ices of the same order.

A square matr ix A is invert ible i f and only if A isnonsingular matr ix.

I f A is a nonsingular matr ix, then i ts inverse

exists which is given by 1 1( )A adj A

A

Consistent system: A system of equat ions is saidto be consistent i f there exist one or more solut ionto the system of equat ion.

I nconsistent system:I f the solut ion to the systemof equat ion does not exist , then i t is termed asinconsistent system.

For the square matr ix A in the matr ix equat ionAX = B A 0, there exists unique solut ion. The system

of equat ion is consistent .

A = 0 and (adj A)B 0 then there exists nosolut ion. The system is inconsistent .

A = 0 and (adj A)B = 0, then system may ormay not be consistent .


1 Mark Questions1. I f for any 2 2 squar e mat r i x A, A(adj A)

=

8 0,

0 8 then wr ite the value of| A| .

[ALL I NDI A 2017]

2. Write A – 1 for

2 5.

1 3A

[DELH I 2011]

3. I f

12 3, then wr i te .

5 2A A

[ALL I NDI A 2015]

4. For what val ues of k , t he syst em of l i nearequat ionsx + y + z = 22x + y – z = 33x + 2y + kz = 4 has a unique solut ion ?

[ALL I NDI A 2016]

2 Marks Question

5. Given

2 3,

4 7A compute A – 1 and show that

2A – 1 = 9I – A.[DELH I 2018]

4 Marks Questions

6. I f

1 2 30 1 4 ,

2 2 1

A

find 1A .

[DELH I 2015]

7. I f A = 2

2 0 1

2 1 3 ,find A 5A 16I .

1 1 0

[ALL I NDI A 2015]

8. I f 3 23

1 0 2

A 0 2 1 andA 6A 7A kI 0

2 0 3

.

Find k.[ALL I NDI A 2016]

CHAPTER 5 : Continuity and Differentiability5.72

[TOPIC 1] Continuity

Summary Definition of Continuity:

Let f is a real valued funct ion and is a subset of

real numbers and a point c l ies in the domain of f,then f is cont inuous at c i f

l imx c

f x f c

When the funct ion f is discontinuous at c, it is called

the point of discont inuity of f.

Also, i f f is defined on [a, b] then cont inuity of a

funct ion f at a means

l imx a

f x f a

And cont inuity of the funct ion f at b means

l imx b

f x f b

Every polynomial funct ion is cont inuous.

Consider two real funct ions f and g which are

cont inuous at c, then sum, difference, product and

quot ient of the two funct ions will also be cont inuous

at x = c.

i .e. (f + g)(x) = f (x) + g (x) is cont inuous at x = c

(f – g)(x) = f (x) – g (x) is cont inuous at x = c

(f . g)(x) = f (x) . g (x)is cont inuous at x = c

Here i f f is a constant funct ion say f(x) = for

some real number , t hen the funct ion ( . g)

defined by (g)(x) = g (x) is also cont inuous.

I f = – 1 then cont inuity of f impl ies cont inuity

of – f.

f xfx

g g x is cont inuous at x = c when g(x) 0

Here, i f f is a constant funct ion say f (x) = for some

real number , then the funct ion g

defined by

xg g x is also cont inuous wherever g(x) 0.

Here are some formulae for limits:

0

l im cos 1x

x

0

sinl im 1x

xx

0

tanl im 1x

xx

1

0

sinl im 1x

xx

1

0

tanl im 1x

xx

0

1l im log , 0

x

ex

aa a

x

0

1l im 1

x

x

ex

0

1l im 1

x

x

ex

0

log 1lim 1e

x

xx

1l im

n nn

x a

x ana

x a

CHAPTER 5 : Continuity and Differentiability 5.77

Summary Different iability:

Consider a real funct ion f and a point c l ies in i tsdomain then the der ivat ive of that funct ion at c isdefined by

0

l imh

f c h f ch

Provided the l imit exists. I t is denoted by f (c) or

df x

dx.

Some rules for algebra of derivatives:

(u + v) = u + v (u – v) = u – v Product rule: (uv) uv + uv

Quot ient rule: 2 , 0

u u v uvv

v v A funct ion which is different iable at a point c is

also cont inuous at that point but the converse isnot t rue.

Chain Rule:

Consider a real value funct ion fwhich is a compositeof u and v.

Let t = u(x) and ,dt dvdx dt

exists, then .df dv dtdx dt dx

Some impor t ant feat ur es of exponent ialfunct ion and logar ithm funct ion are givenbelow

Domain of both the funct ions is set of al l realnumbers.

Range of exponent ial function is set of all posit ivereal numbers and the range of log funct ion isset of al l real numbers.

The point (0, 1) i s always on t he gr aph ofexponent ial function and the point (1, 0) is alwayson the graph of log funct ion.

Both the funct ions are ever increasing.

A relat ion expressed between two var iable x and yin the form x = f(t), y = g(t ) is said to be parametricform with t as a parameter.

By using Chain Rule we find the der ivat ive offunct ion in such form.

[TOPIC 2] Differentiability

dy dy dxdt dx dt

or whenever 0

dydy dxdt

dxdx dtdt

'

'Thus g tdy

dx f t

Second Order Derivative:

If y = f(x)

'dyf x

dx

I f f’(x) is di fferent iable then dy

f xdx

wi l l be

di fferent iated again. The left side wi l l become

d dydx dx and is cal led second order der ivat ive of y

w.r.t . x. Rolle’s Theorem:

Consider a real valued funct ion f defined on theinterval [a, b] such that the funct ion is cont inuouson [a, b], different iable on (a, b) and f(a) = f(b), thenthere exist a point c in (a, b) such that f (c) = 0.

Lagrange’s M ean Value Theorem:

Consider a real valued funct ion f defined on theinterval [a, b] such that the funct ion is cont inuouson [a, b] and different iable on (a, b), then there exists

a point c in (a, b) such that

f b f a

f cb a

.

Derivat ives of some standard funct ions arelisted below:

1n ndx nx

dx

0d

kdx

, k is any constant

log , 0x xe

da a a a

dx

x xde e

dx

1 1

log logloga a

e

dx e

dx x a x

CHAPTER 5 : Continuity and Differentiability5.78

1loge

dx

dx x

sin cosd

x xdx

cos sind

x xdx

2tan secd

x xdx

sec sec tand

x x xdx

2cot cosd

x ec xdx

cosec cosec cotd

x x xdx

1

2

1sin , 1, 1

1

dx x

dx x

1

2

1cos , 1, 1

1

dx x

dx x

12

1tan ,

1d

x xdx x

12

1cot ,

1d

x xdx x

1

2

1sec ,where

1

dx

dx x x

, 1 1,x

1

2

1cosec ,where

1

dx

dx x x

, 1 1,x


2 Marks Question

1. Find dydx

at 1,4

x y

i f 2sin cosy xy K

[DELH I 2017]

4 Marks Questions

2. I f , then findy x b dyx y a

dx

[ALL I NDI A 2017]

3. I f 2 2

1 22 2

1 1tan , 1,

1 1

x xy x

x x

t hen

find dydx

[DELH I 2015]

4. 1 1ye x then show that 22

2d y dy

dxdx

[ALL I NDI A 2017]

5. I f x cos(a + y) = cos y then prove that

2cos a ydydx sin a

Hence show that 2

2d y dy

sin a sin 2 a y 0dxdx

[ALL I NDI A 2016]

6. Find dydx

if 2

1 6x 4 1 4xy sin

5

[ALL I NDI A 2016]

7. Differentiate the following funct ion with respect to

x: loglog x xx x

[DELH I 2013]

8. I f y = log

2 2 ,x x a show that

2

2 22

0d y dy

x a xdxdx

[DELH I 2013]

CHAPTER 6 : Application of Derivative6.96

[TOPIC 1] Rate of Change, Increasing andDecreasing Functions and Approximations

Summary

Rate of Change of Bodies represent ing dydx as a

rate measure:Let us take two var iables x and y that vary withrespect to another var iable says, i .e. i f x = f (s) andy = g(s), then by applying the chain rule, we have

, if .

dydy dxds

dxdx dsds

0

Thus, the rate of change of y with respect to x canbe calculated using the rate of change of y and thatof x both with respect to s.

I ncreasing and decreasing funct ions

A funct ion is said to be increasing when the yvalue increases as the x value increases.

Example:

Y

Y

X X

A funct ion is said to be increasing when the yvalue decreases as the x value increases.

Example:Y

Y

X XO

f is st r ict ly increasing i f

x1 < x2,

f (x1) < f (x2).

Example:Y

Y

X XO

f is st r ict ly decreasing i f x1 < x2, f (x1) < f (x2).

Example:Y

Y

X XO

Let f be cont inuous on [a, b] and different iableon the open interval (a, b). Then:

f is increasing in [a, b] if f (x) > 0 for each x (a, b)

f is decreasing in [a, b] if f (x) < 0 for each x (a, b)

f is a constant funct ion in [a, b] i f f (x) = 0 foreach x (a, b)

Approximat ions

Let the given funct ion be y = f (x) . x denotes asmall increment in x.

The corresponding increment in y is given byy = f (x + x) – f (x)

Different ial of y, denoted by dy is dy

dy xdx

B( + , + )x x y y

A( , )x ydx x =

dy y


[TOPIC 2] Tangents and Normals

Summary A tangent l ine is defined as a st raight l ine that

touches the given funct ion at only one point and i trepresents the instantaneous rate of change offunct ion at the point .

A normal l ine to a point (x, y) on a curve is the l inet hat goes t h r ough t he poi n t (x , y ) and i sperpendicular to the tangent l ine.

tangent line at A

normal line at A

A

• Slope or gradient of a l ine: I f a l ine makes an angle wi th the posi t ive di r ect ion of X axis in ant i -clockwise direct ion, then tan is cal led the slopeor gradient of the l ine.

• I f a tangent l ine to the curve y = f(x) makes anangle with x-axis in the posit ive direct ion, then

slope of the tangent = dy

tandx

• I f slope of the tangent l ine is zero, then tan = 0and so = 0 which means the tangent l ine is parallelto the x-axis. I n this case, t he equat ion of t hetangent at the point is given by (y = y0)

• I f

2

then tan , which means the tangent

l ine is perpendicular to the x-axis, i .e., paral lel tothe y-axis. In this case, the equat ion of the tangentat (x0, y0) is given by (x = x0)

• Equat ion of tangent at (x1, y1) is given by (y – y1)= mT(x – x1), where mT is the slope of the tangent

such that 1 1,

Nx y

dym

dx

• Equat ion of normal at (x1, y1) is given by (y – y1) =mN(x – x1), where mN is the slope of the normal

such that

1 1,

1N

x y

mdydx

• Tangent and normal are perpendicular to eachother, which gives us mT mN = – 1

• I f the slope of two different curves are m1 and m2,then the acute angle between them is given by

2 1

1 2

tan1 .m m

m m

• The slope intercept form of the l ine is y = mx + c,where m is the slope of the given l ine.


4 Marks Questions1. Show that the equat ion of normal at any point

on the curve

x = 3 cos t – cos3 t and y = 3 sin t – sin3 t i s4(y cos3 t – x sin3 t ) = 3 sin 4t

[DELH I 2016]

2. Find the equat ions of the tangent and normal to

the curve 3sinx a and 3cos at4

y a

.

[DELH I 2014]

3. Find the points on the curve 2 2 2 3 0x y xat which the tangents are paral lel to x-axis.

[DELH I 2011]

4. Find the equat ions of the tangent and the normal,

t o t he cu r ve 2 216 9 145x y at t he poi n t

1 1, ,x y where 1 2x and 1 0y .

[DELH I 2018]

CHAPTER 6 : Application of Derivative 6.107

[TOPIC 3] Maxima and Minima

Summary• The maximum value attained by a function is called

maxima and the minimum value at tained by thefunct ion is known as minima.

• Consider y = f (x) be a well-defined funct ion on aninterval I , then

f is said to have a maximum value in I , i f thereexist a point c in I such that f (c) > f (x), x I .The value cor responding to f(c) i s cal led asmaximum value of x in I and the point c is themaximum value.

f is said to have a minimum value in I , i f thereexist a point c in I such that f (c) < f (x), x I .The value cor responding to f(c) i s cal led asminimum value of x in I and the point c is theminimum value.

f is said to have an ext reme value in I , i f thereexist a point c i n I such t hat f (c) i s ei t hera maximum value or a minimum value. Thevalue cor responding to f(c) is cal led as ext remevalue of x in I and the point c is the ext remepoint.

• Let f be a funct ion defined on an open interval I .Suppose c ? I be any point . I f f has a local maximaor a local minima at x = c, then either f (c) = 0 or fis not different iable at c.

First Derivative Test

Let f be a funct ion defined on an open interval I .

Let f be cont inuous at a cr i t ical point c in I . Then

• I f f (x) changes sign from posi t ive to negat iveas x i ncr eases t hr ough c, i .e., i f f (x) > 0 atevery point sufficient ly close to and to the leftof c, and f (x) < 0at every point sufficient ly closeto and to the r ight of c, then c is a point of localmaxima.

• I f f (x) changes sign from negat ive to posi t iveas x increases through c, i .e., i f f (x) < 0 at everypoint suff icient ly close to and to t he left of c,and f (x) > 0 at every point sufficient ly close toand to the r ight of c, then c is a point of localminima.

• I f f (x) does not change sign as x increases throughc, then c is neither a point of local maxima nor apoint of local minima. I n fact , such a point is cal ledpoint of inflect ion.

point oflocal maxima point of non differentiability

and point of local maxima

point of non differentiability and point of local minima

pointof local minima

C1 C2 C3 C4

f (x)>0

f (c )=0 1 f (x)<0

f (c )=

0

2f (

x)>0

Y

Y

X XO

Second Derivative Test

Let f be a funct ion defined on an interval I and c I .Let f be twice different iable at c. Then

• x = c is a point of local maxima i f f (c) = 0 andf (c) < 0. The value f (c) is local maximum value of f.

• x = c is a point of local minima if f (c) = 0 and f (c) > 0In this case, f (c) is local minimum value of f.

• The test fai ls i f f (c) = 0 and f (c) = 0. I n this case,we go back to the fi rst der ivat ive test and findwhether c is a point of local maxima, local minimaor a point of inflexion.

• M aximum and M inimum values of a functionin a closed interval

Let f be a cont inuous funct ion on an int er valI = [a, b]. Then f has the absolute maximum valueand f at tains i t at least once in I . Also, f has theabsolute minimum value and attains it at least oncein I .

Let f be a different iable funct ion on a closed intervalI and let c be any inter ior point of I . Then

f (c) = 0 i f f at tains i ts absolute maximum valueat c.

f (c) = 0 i f f at tains i ts absolute minimum valueat c.

I n view of the above results, we have the fol lowingworking rule for finding absolute maximum and/or absolute minimum values of a funct ion in a givenclosed interval [a, b].


PREVIOUS YEARS’EXAMINATION QUESTIONS

TOPIC 34 Marks Questions

1. I f t he funct ion 3 2 2( ) 2 – 9 12 1f x x mx m x ,

where m > 0 at tains i ts maximum and minimum

at p and q respect ively such that 2p q , then

find the value of m.

[ALL I NDI A 2015]

2. A rectangular sheet of t in 45 cm by 24 cm is to bemade into a box without top, by cutt ing off squarefrom each corner and folding up the flaps. Whatshould be the side of the square to be cut off so thatthe volume of the box is the maximum possible?

[ALL I NDI A 2011]

6 Marks Questions3. Show that the surface area of a closed cuboid with

square base and given volume is minimum, whenit is a cube.

[ALL I NDI A 2017]

4. The volume of a cube is increasing at the rate of9 cm3/s. How fast is i ts sur face area increasingwhen the length of an edge is 10 cm ?

[ALL I NDI A 2017]

5. I f the sum of lengths of the hypotenuse and aside of a r ight angles t r iangle is given, show thatthe area of the t r iangle is maximum, when the

angle between them is .3

[DELH I 2017]

• Working Rule

Find al l cr i t ical points of f in the interval, i .e.,find points x where either f (x) = 0 or f is notdifferent iable.

Take the end points of the interval.

At al l t hese point s (l i sted in Step 1 and 2),calculate the values of f

6. Show that the semiver t ical angle of the cone ofthe maximum volume and of given slant height is

1 1cos

3

[DELH I 2014]

7. Show t hat t he al t i t ude of t he r i ght ci r cularcone of maximum volume that can be inscr ibed

in a sphere of r adius r is 4

.3r

Also find the

maximum volume in t erms of volume of t hesphere.

[ALL I NDI A 2014]

8. Show that of al l the rectangles inscr ibed in agiven fixed circle, the square has the maximumarea,

[DELH I 2011]

9. Show t hat t he hei gh t of a cl osed r i gh tci r cu l ar cy l i nder of gi ven su r face andmaximum volume, i s equal t o t he diamet erof i ts base.

[DELH I ]

10. Show t hat semi -ver t i cal angl e of a cone ofmaximum volume and gi ven slant height i s

1 1cos

3.

[ALL I NDI A 2016]

11. Show that the alt i tude of the r ight circular coneof maximum volume that can be inscr ibed in asphere of radius r is 4r /3, Also show that themaximum volume of t he cone is 8/27 of thevolume of the sphere.

[DELH I 2014]

I dent i fy the maximum and minimum values of fout of the values calculated in

This maximum val ue wi l l be t he absolut emaximum (greatest ) value off and the minimumvalue will be the absolute minimum (least) valueof f.

CHAPTER 7 : Integrals7.120

[TOPIC 1] Indefinite Integrals

Summary Integrat ion is the inverse of different iat ion. Instead

of different iat ing a funct ion, we wil l be given theder ivat ive of a funct ion and we would be asked tofind i ts pr imit ive funct ion. Such a process is cal ledintegrat ion or ant i di fferent iat ion.

I f ( ) ( )d

F x f xdx

. Then we write ( ) ( )f x dx F x C .

These integrals are cal led indefini te integrals orgeneral integrals and C is cal led the constant ofintegrat ion.

The integral of a funct ion are unique upto anaddi t i ve const ant , i .e. any t wo int egr als of afunct ion differ by a constant .

When a polynomial funct ion P is integrated, theresult is a polynomial whose degree is one morethan that of P.

Geometr ical ly, the indefini te integral of a funct ionrepresents a family of curves placed parallel to eachother having paral lel tangents at the points ofintersect ion of the curves of the family with thel ines perpendicular to the axisrepresent ing thevar iable of integrat ion.

Two indefini te integrals with the same der ivat ivelead to the same family of curves and so they areequivalent .

Some pr oper t i es of i ndef ini t e i nt egr al ar e asfol lows:

( ) ( ) ( ) ( )f x g x dx f x dx g x dx

For any real number a, ( )af x dx a f x dx

Proper t ies (i) and (i i ) can be general ised to afini te number of funct ions f1, f2, f3,...fn and thereal numbers, a1, a2, a3,...an giving

1 1 ( )dxa f x 2 2 ( )dx ... a ( )n na f x f x dx

1 1 2 2(x)dx a ( )dx ...a ( )n na f f x f x dx .

Some basic integrals are as fol lows:

1

,n 11

nn x

x dx Cn

dx x C

cos sinxdx x C

sin cosxdx x C

2sec tanxdx x C

2cosec cotxdx x C

sec tan secx xdx x C

cosec cot cosec x x dx x C

1

2sin

1

dxx C

x

1

2cos

1

dxx C

x

1

2 tan1

dxx C

x

1

2 cot1

dxx C

x

1

2sec

1

dxx C

x x

1

2cosec

1

dxx C

x x

x xe dx e C

1

logdx x Cx

log

xx a

a dx Ca

I ntegrat ion by subst itut ion:

The method in which we change the var iable tosome ot her var i abl e i s cal l ed t he met hod ofsubst i tut ion. I t often reduces an integral to one ofthe fundamental integrals.

The integral f(x)dx can be subst i tuted into anotherform by changing the independent var iable x to tby subst i tut ing x = g(t ).

CHAPTER 7 : Integrals 7.121

Consider, A = f(x)dx

Put x = g(t ), therefore dxdt

g t .

dx = g(t )dt

Thus, A f x dx f g t dtt g

Using subst i tut ion method we obtain the fol lowingstandard integrals.

tan log secxdx x C

cot log sinxdx x C

sec log sec tanxdx x x C

cosec log cosec x cotxdx x C

I ntegrals of some par t icular funct ions areas follows:

2 2

1log

2dx x a

Ca x ax a

2 2

1log

2dx a x

Ca a xa x

1

2 2

1tan

dx xC

a ax a

2 2

2 2log

dxx x a C

x a

1

2 2sin

dx xC

aa x

2 2

2 2log

dxx x a C

x a

I ntegrat ion by Part ial Fract ions:

A rat ional funct ion is defined as the rat io of two

polynomials in the form ( )( )

P xQ x

, where P(x) and

Q(x) are polynomials in x and Q(x) 0. I f ( )( )

P xQ x is

improper, then 1 ( )( )( )

( ) ( )P xP x

T xQ x Q x where T(x) is

a polynomial in x and 1 ( )( )

P xQ x is a proper rat ional

funct ion. Assume we want to evaluate ( )( )

P xdx

Q x ,

where ( )( )

P xQ x

is a proper rat ional funct ion. I t is

possible to wr ite the integrand as a sum of simpler

rat ional funct ions by part ial fract ion decomposit ionas fol lows:

,px q A B

a bx a x b x a x b

2 2

px q A Bx ax a x a

2px qx r A B Cx a x b x c x a x b x c

2

2 2

px qx r A B Cx a x bx a x b x a

2

22,

px qx r A Bx Cx a x bx cx a x bx c

wher e

x2 + bx + c cannot be factor ized fur ther.

I ntegrat ion by parts:

I f f (x) and g(x) ar e t he t wo funct i ons t hen,

'( ) ( ) ( ) ( ) ( ) ( )f x g x dx f x g x dx f x g x dx dx ,

where f(x) is the first funct ion and g(x) is the secondfunct i on . I t can be st at ed as fol l ows: “ T heintegrat ion of the product of two funct ions= (F irst funct ion) x (integral of the secondfunct ion) – I nt egr al of [ (di ffer ent ialcoefficient of the first funct ion) x (integralof the second funct ion)]”

I ntegral of the type 'x xe f x f x dx e f x C

Some special types of integrals are as fol lows:

2

2 2 2 2 2 2log2 2x a

x a dx x a x x a C

2

2 2 2 2 2 2log2 2x a

x a dx x a x x a C


2

2 2 2 2 1sin2 2x a x

a x dx a x Ca

I n t egr al s of t he t ypes 2

dxax bx c

or

2

dx

ax bx c can be t r ansfor med i n t o

st andar d for m by expr essi ng 2ax bx c

2 b c

a x xa a

2 2

22 4b c b

a xa a a

I n t egr al s of t he t ypes 2

px qdxax bx c

or

2

px qdx

ax bx c can be t r ansfor med i n t o

st andar d for m by expr essi ng px q

2dA ax bx c B

dx 2A ax b B ,

where A andB are determined by compar ingcoefficients on both sides.


1 Mark Questions

1. Write the ant i der ivat ive of 1

3 xx

.

[DELH I 2014]

2. Evaluate 1 .x xdx[ALL I NDI A 2012]

3. Write the value of 2 16

dx

x .

[DELH I 2011]

4. Write the value of sec sec tanx x x[DELH I 2011]

5. Evaluate 21 log x

dxx

[ALL I NDI A 2011]

6. Find : 2 2sin cosx x

dxsinxcosx

[ALL I NDI A 2017]

2 Marks Questions

7. Find 2 4 8

dx

x x

[DELH I 2017]

8. Find : 25 8

dx

x x

[ALL I NDI A 2017]

9.2

2cos2 2sin

cos

x xd

xx

[DELH I 2018]

4 Marks Questions

10. Evaluate:

sin

sin

x adx

x a

[DELH I 2013]

11. I ntegrate the fol lowing w.r.t . x: 2

2

3 1

1

x x

x

.

[DELH I 2015]

12. Evaluate: 2

2 24 9

xdx

x x

[DELH I 2013]

13. Find 22 2

2

1 2

xdx

x x

[DELH I 2017]


[TOPIC 2] Properties of a Definite Integrals and Limitof a Sum

Summary Definite I ntegrals:

A def i n i t e i n t egr al i s denot ed by b

af ( x )dx ,

wher e a i s cal l ed t he l ower l i m i t of t heintegral and b i s cal led the upper l imi t of theintegral.

Definite integral as the limit of a sum:

The definite integral b

af ( x )dx is the area bounded

by the curve y = f(x), the ordinates x = a, x = b andthe x-axis.This can be mathemat ical ly defined asfol lowing:

0

1b

a hf x dx b a l im f a f a h

n

1... f a n h

Where, 0 as

b ah n

n ( )b

a

f x dx is defined

as the area funct ion where the area of the regionis bounded by the curve y = f(x), a x b, thex – axis and the ordinates x = a and x = b. Let x be

a given point in [a, b]. Then ( )x

a

f x dx represents

the Area function A(x).

F i r st fundament al t heor em of int egr alcalculus:

Let f be a cont inuous funct ion on the closed interval

[a, b] and let ( ) for al lb

a

A x f x dx x a be the

area funct ion. Then A(x) = f(x), for al l x [a, b].

Second fundament al t heor em of int egr alcalculus:

Let f be cont inuous funct ion defined on the closedinterval [a, b] and F be an ant i der ivat ive of f. Then

( )b

b

aa

f x dx F x F b F a .

Properties of Definite I ntegrals are as fol lows:

0 : ( ) ( )b b

a a

P f x dx f t dt

1 .b a

a b

P f x dx f x dx

I n par t icular ( ) 0a

a

f x dx

2 :b c b

a a c

P f x dx f x dx f x dx

3 :b b

a a

P f x dx f a b x dx

40 0

:a a

P f x dx f a x dx

2

50 0 0

: 2a a a

P f x dx f x dx f a x dx

2

60 0

: ( ) 2 ( ) , i f (2 ) ( )a a

P f x dx f x dx a x f x

and 0, i f (2a – x) = – f(x).

7 :P

0

( ) , i f f is an even funct ion, 2i.e., f ( ) ( )(2 ) ( )

( )0, i f f is an odd funct ion,

i .e., i f ( ) ( )

a

a

a

f x dxx f x a x f x

f x dx

f x f x

CHAPTER 8 : 8.152

Summary Area under simple curves

Consider that a curve y = f(x), the l ine x = a,x = b and x-axis col lect ively acquires an areaand the area under the curve is considered ascomposed of large number of ver t ical thin st r ips.

Now assume that there is an arbitrary str ip withheight y and width dx.

Then dA which represents area of elementaryst r ip = ydx, where y = f(x).

Total area A of the region between the curvey = f(x), x = a, x = b and x-axis is equal to thesum of areas of all elementary vertical thin str ipsacross the region PQRS.

which is given by, b b

a aA y dx f (x)dx

dx

Y

R

X

y f x= ( )

x a=x b=

S

y

QPOY

X

Elementary Area: The area which is locatedat an arbit rary posit ion within the region whichis specified by some value of x between a and b

Now consider the area A of the region which isbounded by the curve x = g(y), the l ines y = c,y = d and y-axis.

Total area A of the region between the curvex = g(y), y = c, y = d and y-axis is equal to thesum of areas of al l elementary hor izontal thinst r ips.

I n this case the area A is given by

d d

c cA x dy g(y)dy

X

y d =

xdy

y = c

x = g y( )

Y

XOY

I f the curve is posit ioned below x-axis, whichi s f (x) < 0 f r om x = a t o x = b, t hen t henumer i cal val ue of t he ar ea wh i ch i sbounded by t he cur ve y = f (x), x-axi s andthe ordinates x = a, x = b wi l l come out t obe negat ive. But , i f t he numer ical value oft he ar ea i s t o be t aken int o consider at ion,then is given by:

b

aA f (x) dx

Xx b=x a=

y f x= ( )

O

Y

Y

X

There is a possibi l i tythat some por t ion of thecurve is located above x-axis and some por t ionof i t is located below x-axis.

Let suppose A1 is the area below x-axis and A2isthe area above x-axis. Now, the area of the regionwhich is bounded by the curve y = f(x), x = a,x = b and x-axis can be given by A = A1 + A2.

Application of Integrals

CHAPTER 8 : Application of Integrals 8.153

Y

x a= O

A1

A2

X X

x b=

Y

Area of the region bounded by a curve and a

line

Area of the region bounded by a l ine and a curve

is used to find the area bounded by a l ine and a

parabola, a l ine and an el l ipse, a l ine and a circle

etc. The standard equat ion wil l be used for these

ment ioned curves.

Area of the region can be calculated by taking

the sum of t he ar ea of ei t her hor izont al or

ver t ical elementary str ips but ver t ical st r ips are

most ly prefer red.

Area between two curves

Assume that there are two curves, y = f(x) and

y = g(x), where f(x) g(x) in [a, b]. The ordinates

x = a and x = b give the point of intersect ion of

these two curves. Suppose that these curves

intersect at f(x) with width dx.

Consider an elementary ver t ical st r ip of height

y, where y = f(x).

dA = y dx

Now the area is given by,

b

aA f (x) g(x) dx

b b

a af (x) dx g(x)dx

which can be stated as,

A = Area bounded by the curve {y = f(x)}

– Area bounded by the curve {y = g(x)}

where f(x) > g(x).

y f x g x = ( ) – ( )

y

y = g x( )x = b

x = a

dx

X

y f x = ( )Y

XOY

X

I n other case i f t he two curves y = f(x) and

y = g(x) where f(x) g(x) in [a, c] and f(x) g(x) in

[c, b] with a condit ion that a < c < b, intersect at

x = a, x = c and x = b, then the area bounded by

the curves is given by:

c b

a cA f (x) g(x) dx g(x) f(x) dx

which is stated as:

Total area = Area of the region ACBDA

+ Area of the region BPRQB

y f x = ( )y g x = ( )

y g x = ( ) y f x = ( )

x a= x c= x b=

Q

RP

BC

X

A

D

Y

OX

Y

CHAPTER 9 : Differential Equations9.166


1 Mark Questions

1. Find the different ial equat ion represent ing thecurve y = cx + c2.

[ALL I NDI A 2015]

2. Write the different ial equat ions represent ing thefamily of curves y = mx, where m is an arbit raryconstant .

[ALL I NDI A 2013]

3. Write the degree of the different ial equat ion

2 423

20.

d y dyx x

dxdx

[DELH I 2013]

[TOPIC 1] Formation of Differential Equations

Summary Differential Equation: Different ial equat ion is

an equat ion which involves der i vat i ve of t hedependent var iable with respect to independent

var iabl e. H er e dy

x y 0dx

i s t he example of

different ial equat ion.

General notat ions for der ivat ives are:

2 3

2 3

dy d y d yy , y , y

dx dx dx

or n

nn

d yy

dx

Order of a differential equation: The order oft he h i ghest or der der i vat i ve i n any gi vendi fferent ial equat ion is cal led the order of t hedifferent ial equat ion.

Example: sin 0dy

xdx

has the order 2.

D egr ee of a di f fer ent ial equat ion: I f adi f fer en t i al equat i on i nvol ves a pol ynomi alequat ion in i ts der ivat ive, then i ts degree can bedefined as the highest power of the highest orderder ivat ive in i t .

Example: sin 0 dy

xdx

has the degree 1.

Degree (i f defined) and order are always posit ive.

4. Find the different ial equat ion represent ing the

family of curves ,A

Br

where A and B are

arbit rary constants

[DELH I 2015]

2 Marks Question5. Find the different ial equat ion represent ing the

family of curves 5bxy ae , where a and b are

arbit rary constants.

[DELH I 2018]

4 Marks Questions6. Find the different ial equat ion for al l the st raight

l ines, which are at a uni t dist ance fr om the

or igin.

[ALL I NDI A 2015]

CHAPTER 9 : Differential Equations9.168

[TOPIC 2] Solution of Different Types of DifferentialEquations

Solut ion of a different ial equat ion: Solut ion

of a different ial equat ion is a funct ion sat isfying

that different ial equat ion.

Gener al Solut ion: The sol u t i on havi ng

the ar bi t r ar y constant s (equal t o t he or der

of t he di f f er en t i al equat i on ) i s cal l ed a

general or pr imit ive solut ion of the different ial

equat ion.

Par t icular Solut ion: The sol u t i on wh i ch

does not have the arbit rary constants is cal led

a par t i cu l ar sol u t i on . I t i s acqu i r ed by

subst i t u t i ng t he par t i cu l ar val ues i n t he

arbit rary constants.

I f the general solut ion of any different ial equat ion

is given, then the funct ion is to be different iated

successively (as many t imes as the total number

of ar bi t r ar y const ant s) i n or der t o for m t he

di f fer ent i al equat ion and t hen t he ar bi t r ar y

constants are el iminated.

A di fferent ial equat ion which can be separated

completely such that the terms which contains x

can be wr it ten with dx and that of containing y

with dy, can be solved with the help of var iable

separable method. The solut ion of such equat ions

are of the form f(x)dx = g(x)dx + C, where C is an

arbit rary constant .

H omogeneous D i ffer ent i al E quat ion: A

different ial equat ion of the form dy

f (x, y)dx

is

cal led as homogeneous different ial equat ion i f f(x, y) is a homogeneous funct ion of degree 0.

To solve a homogeneous equat ion of t he form

dy y

f (x,y) gdx x

, subst i t ut ions are made as

y

vx

or y = vx and then general solut ion is found

out by solving dy dv

v xdx dx

.

To solve a homogeneous equat ion of t he form

dx

f (x,y)dy

, subst i tut ions are made as x

vy

or

x = vy and then general solut ion is found out by

wr it ing dx dv

v ydy dy

.

Linear Different ial Equat ion: A di f ferent ial

equat ion which can be expressed as dy

Py Qdx

,

where P and Q are the constants or funct ions of xonly, is called a l inear different ial equat ions of fi rstorder.

To solve a l inear different ial equat ion, i t is fi rst

wr it ten as dy

Py Qdx

, then the integrat ing factor

is found as Pdx

I .F. e . After that , the solut ion is

given by y(I.F.) (Q I.F.)dx C .

I f the l inear different ial equat ion is of the form

dx

Px Qdy

(P and Q are constants or funct ions

of y only), then Pdy

I .F. e and the solut ion is given

by x(I.F.) (Q I.F.)dy C .

CHAPTER 10 : Vector Algebra10.184

[TOPIC 1] Algebra of Vectors

Summary A vector quantity has both magnitude and direction

where the magnitude is a distance between theinitial and terminal point of the vector. Let'sassume a vector starts at a point A and ends at apoint B. Therefore the magnitude of the vector is

denoted by AB .

OA r xi y j zk is the position vector of any

point A(x, y, z) having a magnitude equal to

2 2 2x y z . Where O is the origin (0, 0, 0) and

P is any point in the space.

The angles are known as the direction angleswhich are made by the position vector and thepositive x, y, z – axes respectively and their cosinevalues (cos, cos, cos) are known as directioncosines, denoted by l, m, n respectively.

The projections of a vector along the respectiveaxes are represented by the direction ratios whichare the scalar components of the vector. They aredenoted by a, b, c respectively.

The direction cosines, direction ratios, andmagnitude of a vector are related as:

, ,a b c

l m nr r r

In general, l2 + m2 + n2 = 1 but a2 + b2 + c2 1.

Zero vector (also known as a null vector) is

symbolized by 0

. Its initial and terminal points

coincide.

A unit vector has a magnitude equal to 1 and is

denoted by a .

If two or more than two vectors have the sameinitial points, they are called as co-initial vectors.

The vectors which are parallel to the same lineare known as collinear vectors.

The vectors having equal magnitude and samedirection are called equal vectors.

A vector having the same magnitude as the givenvector but opposite direction is known as thenegative of the given vector.

Triangle law of vector addition: Let's say thatA, B, and C are the vertices of a triangle then

C

A B

AC AB BC

Parallelogram law of vector addition: If twovectors are represented by the two adjacent sidesof a parallelogram, then their sum is representedby the diagonal of that parallelogram through theircommon point. For example, the

C

AO

B

bb + a

ba + b

a

a

OC OA OB

OC a b

Vector addition is commutative as well as associativein nature and also has zero vector as an additiveidentity.

The multiplication of any vector a by a scalar is

denoted by a and has the same direction as the

original vector if is positive and opposite direction

if is negative. Its magnitude is a a .

CHAPTER 10 : 10.185

The unit vector of any vector a in its direction is

written as 1 a aa

The unit vectors along the positive x, y, z axes are

denoted by , , i j k respectively.

Component form of a vector: The component

form of any vector is r xi y j zk where x, y, z

are called as the scalar components and , ,z xi y j k

as the vector components of r . x, y, z is also called

the rectangular components.

If two vectors are in their component form as

1 2 3 a a i a j a k and 1 2 3

b b i b j b k , then

The sum of the vectors a and

b is given by

1 1 2 2 3 3 a b a b i a b j a b k .

The difference of the vectors a and

b is given

by 1 1 2 2 3 3 a b a b i a b j a b k

The vectors a and

b are equal if

1 1 2 2 3 3, anda b a b a b .

The multiplication of a vector a by scalar is

given by 1 2 3 a a i a j a k .

Vector joining two points: The magnitude of a

vector 1 2A A joining two points 1 1 1 1( , , )

A x y z and

2 2 2 2( , , )A x y z is

2 2 21 2 2 1 2 1 2 1( ) ( ) (z )A A x x y y z

Section Formula: The position vector of a pointC dividing the line segment joining two points A

and B (having position vectors , a b respectively) in

the ratio of m : n

Internally:

mb nar

m n

Externally:

mb nar

m n

If C is the midpoint of A and B, then

2

a br


1 Mark Questions

1. I f ˆ ˆˆ ˆ ˆ ˆ2 and 3a xi j zk b i yj k ar e t wo

equal vectors, then wr i te the value of x + y + z

[DELH I 2013]

2. I f a un i t vect or a mak es angl es 3

w i t h

ˆ, with4

i j

and an acute angle ˆ, 0,1with k

then find the value of .[DELH I 2013]

3. Find the Car tesian equat ion of the l ine whichpasses through the point (– 2, 4, – 5)and is the

paral lel to the l ine 3 4 8

.3 5 6

x y z

[DELH I 2013]

4. I f a l ine makes angles 90°.60°and with x, yand z-axis repect ively, where is acute, thenfind .

[DELH I 2015]

5. I f a l ine makes angle 90 and 60 respect ivelywi th the posit ive direct ion of x and y axes, findt he angle which i t makes wi t h t he posi t i vedirect ion of z – axis.

[DELH I 2017]

6. Find the vector equat ion of t he l ine passingthrough the point A 1,2, 1 and paral lel to thel ine 5 25 14 7 35 . x y z

[DELH I 2017]

Vector Algebra

CHAPTER 10 : Vector Algebra 10.189

[TOPIC 2] Product of Two Vectors, Scalar TripleProduct

Summary

Scalar (or dot) product of two vectors a

and b : The scalar product of two vectors having

an angle between them is denoted by a b and is

defined by

cos a b a b

cos

a b

a b

or

1cos

a b

a b

If 0 a b a b

Properties of scalar product:

Let , and a b c be any three vectors then

a b c a b a c

Let and a b be any two vectors, and be any

scalar. Then a b a b a b

Projection of a vector a on another vector

b is

given as

a b

a.

Projection of a vector b on another vector

a is

given as

a b

b.

Vector (or cross) product of two vectors a

and b : The vector product of two vectors having

an angle between them is denoted by a b and

is defined by:

sin a b a b n

sin

a b

a b

If 0

a b a b

If the vectors are in their component form as

1 2 3 a a i a j a k and 1 2 3

b b i b j b k , then

their cross product is given by

1 2 3

1 2 3

i j ka b a a a

b b b

The dot product is given by

1 1 2 2 3 3 a b a b a b a b

Properties of vector product:

Let , and a b c be any three vectors then

a b c a b a c .

Let and a b be any two vectors, and be any

scalar. Then a b a b a b

Scalar triple product of three vectors [A, B, C] is

given by . A B C which is the volume of a

parallelepiped whose sides are given by vectors

, A B and

C .

CHAPTER 11 : Three Dimensional Geometry11.202

[TOPIC 1] Direction Cosines and Lines

Summary Direction Cosines:

These are the cosines of the angles made by theline with the positive directions of the coordinateaxes.

P

Z

L

X

zr

y YOx

The direction cosines of line joining the points A(x1,y1, z1 and B(x2, y2, z2) are

2 1 2 1 2 1, ,

x x y y z zAB AB AB

where

2 1 22 2

2 12

1x x y zB zA y

Assume the direction cosines of the line are l, mand n and that the line is passing through the pointA(x1, y1, z1), then the equation of the line is

1 1 1x x y y z z

l m n

Also, l2 + m2 + n2 = 1

Direction Cosines:Direction ratios are any three numbers which areproportional to the direction cosines.

Let the direction ratios be a, b and c, then

2 2 2 2 2 2

,a b

l ma b c a b c

and

2 2 2

cn

a b c

Lines: Equation of the line which passes through two

given points:Assume that the position vectors of A(x1, y1, z1)

and B(x2, y2, z2) are a and

b respectively.

( , , )x y z

11

1

( , , )x y z

22

2

( , , )x y z

Z

P

B

A

O

X

abr

Y

Then, the vector equation of the line is

,r a k b a k

The equation of the line in Cartesian form is

1 1 1

2 1 2 1 2 1

x x y y z zx x y y z z

Equation of the line which passes through a givenpoint having a given direction

O

X

Y

A

Z

P l

ra

b

The vector equation of the line is r a b

The equation of the line in Cartesian formwhen it passes through A(x1, y1, z1) is

1 1 1x x y y z z

a b c where the direction

ratios are a, b and c.

CHAPTER 11 : Three Dimensional Geometry 11.203

The angle between the lines

^ ^ ^1 1 1r a i b j c k

and

^ ^ ^2 2 2r a i b j c k is given by

1 2 1 2 1 22 2 2 2 2 2

1 1 1 2 2 2

cosa a b b c c

a b c a b c

The shortest distance between the lines

1 1r a b and

2 2r a b is given by

1 2 2 1

1 2

b b a a

b b

The shortest distance between the lines

1 1 1

1 1 1

x x y y z za b c and

2 2 2

2 2 2

x x y y z za b c is given by

2 1 2 1 2 1

1 1 1

2 2 22 2 2

1 2 2 1 1 2 2 1 1 2 2 1

x x y y z za b ca b c

b c b c c a c a a b a b


1 Mark Questions1. I f t he Car t esi an equat i ons of a l i ne ar e

3 4 2 6,

5 7 4x y z Wr ite the vector equat ion

for the l ine. [ALL I NDI A 2014]

2. I f a l ine has direct ion rat ios 2, -1, -2, then whatare i ts direct ion cosines?

[ALL I NDI A 2018]

3. Wr i t e t he di r ect i on cosi nes of t he vect or

ˆˆ ˆ2 5 .i j k

[DELH I 2011]

4. What ar e t he di r ect i on cosi nes of a l i ne,which makes equal angles with the coordinateaxes?

[ALL I NDI A 2014]

2 Marks Question5. The x-coordinate of a point l ies on the line joining

the points P(2, 2, 1) and Q (5, 1, – 2) is 4. Find i tsz-coordinate.

[ALL I NDI A 2017]

4 Marks Questions

6. Show t hat t he l i nes1 3 5

3 5 7x y z

and

2 4 61 3 5

x y z inter sect . Also f ind thei r

point of intersect ion.

[DELH I 2014]

7. Find the shor test distance between the l ines:

2 ˆ ˆˆ ˆ 2ˆ3 3 ˆr i j k i j k

4 5 ˆ ˆˆ ˆ ˆ6 2ˆ 3r i j k i j k

[ALL I NDI A 2011]

8. The scalar product of the vector ˆˆ ˆ a i j k

wi t h a uni t vect or al ong t he sum of vect or

2 4 ˆˆ ˆ 5b i j k

and 2 3 ˆˆc i j k is equal

to one. Find the value of and hence find the

unit vector along .b c

[ALL I NDI A 2014]

9. Find the coordinates of the foot of perpendiculardrawn fr om the point A(– 1, 8, 4) t o t he l inejoining the points B(0, – 1, 3) and C(2, – 3, – 1).Hence find the image of the point A in the l ineBC.

[ALL I NDI A 2016]

CHAPTER 11 : Three Dimensional Geometry 11.209

Summary Plane

A surface so that when the two points are taken onit, the line segment lies joining the two points lieson the surface is called a plane.

The equation of the plane is ^r n d where ^n

is the unit vector normal to plane of origin.The equation of the plane in normal form islx + my + nz = d where l, m, n are directioncosines.

Equation of a plane perpendicular to a givenvector

O YP( , , )x y z

X

Z

Aa

r

The equation of a plane through a point whose

position vector is a and perpendicular to the vector

N is

0r a N

Equation of a plane perpendicular to a givenvector and passing through a given point isA(x – x1) + B(y – y1) + C(z – z1) = 0

Equation of a plane passing through threenon collinear points

Let the non collinear points be R(x1, y1, z1),

S(x1, y2, z2), T(x3, y3, z3) and r be the position

vector.

[TOPIC 2] Plane

Z

Y

X

RP

O

S T

(RS RT)

br a c

Equation of a plane passing through three noncollinear points is

0r a b a c a

In Cartesian plane,

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

0x x y y z zx x y y z zx x y y z z

The equation of the plane in the intercept form is

1x y za b c

where a, b, c are x, y, z intercepts

respectively.

The plane passing through intersection of twogiven lines

It has the equation

1 2 1 2r n n d d

In Cartesian system,

(A1x + B1y + C1z – d1) + l(A2x + B2y + C2z – d2)

= 0

CHAPTER 11 : Three Dimensional Geometry11.210

Angle between two planes

The angle between the planes is given by the anglebetween their normals.

angle between the normals= [90 – (90 – )]Plane 1

the angle betweenthe planes

Plane 290 –

2n

1n

Let the angle between the planes be .

1 2

1 2

cosn n

n n

where 1 2,n n are normal to the

planes.

In Cartesian form, A1x + B1y + C1z + D1 = 0 andA2x + B2y + C2z + D2 = 0

1 2 1 2 1 2

2 2 2 2 2 21 1 2 21 2

cosA A B B C C

A B C A B C

The distance between a plane Ax + By + Cz + Dand the point (x1, y1, z1) is given by

1 1 1 12 2 2

A x B y C z D

A B C

The angle between the line r a b and the

plane

^r n d is

^

^sin

b n

b n


1 Mark Questions1. Find the distance between the planes 2x – y + 2z = 5

and 5x – 2.5y + 5z = 20.[ALL I NDI A 2017]

2. Write the sum of intercepts cut off by the plane

ˆ. 0ˆ2 5r i j k on the three axes.

[ALL I NDI A 2016]

3. F i nd t he acu t e angl e bet ween t he pl ane5x – 4y + 7z – 13 = 0 and the y-axis

[ALL I NDI A 2015]

4. Find the length of the perpendicular drawn fromor igin to the 2x – 3y + 6z + 21 = 0 plane .

[ALL I NDI A 2013]

5. Find the vector equat ion of a plane which is at adistance of 5 units from the or igin and i ts normal

vector ˆˆ ˆ2 3 6i j k [DELH I 2016]

6. Find , i f the vectors 3a i j k , 2b i j k

and 3c j k

are coplanar..

[DELH I 2015]

4 Marks Questions7. Find the distance between the point 1, 5, 10

and t he poi n t of i n t er sect i on of t he l i ne

2 1 23 4 12

x y z and the plane 5.x y z

[DELH I 2015]

8. Show that the vectors , and ca b are coplanar i f

, cand ca b b a are coplanar..

[DELH I 2016]

9. Find the equat ion of the plane determined by thepoints A(3, – 1, 2), B(5, 2, 4) and C(– 1, – 1, 6) andhence find the distance between the plane andthe point P(6, 5, 9).

[ALL I NDI A 2012]

10. Show that the four points A, B, C and D with

posit ion vectors ˆˆ ˆ4 5 , ,ˆi j k j k ˆˆ ˆ3 9 4i j k

and 4( )ˆˆ ˆi j k are respect ively coplanar..

[ALL I NDI A 2014]

CHAPTER 12 : Linear Programming12.224

Summary Linear Programming

Linear programming is a method which providesthe optimization (maximization or minimization)of a linear function composed of certain variablessubject to the number of constraints.

Applications of Linear Programming

Used in finding highest margin, maximum profit,minimum cost etc.

Used in industry, commerce, managementscience etc.

Linear Programming Problem (LPP)

Linear Programming problem is a type of problemin which a linear function z is maximized orminimized on certain conditions that aredetermined by a set of linear inequalities with non-negative variables.

Mathematical Formulation of LPP

Optimal value: Maximum or Minimum valueof a linear function

Objective Function: The function which is tobe optimized (maximized/minimized).

Linear objective function: Z = ax + by is alinear function form, where a, b are constants,which has to be maximized or minimized is calleda linear objective function.

For example- Z = 340x + 60y, where variables xand y are called decision variables.

Constraints: The limitations as disparities on thefactors of a LPP are called constraints. Theconditions x 0, y 0 are called non-negativerestrictions.

'Linear' states that all mathematical relations usedin the problem are linear relations. Programmingrefers to the method of determining a particularprogram or plan of action.

Mathematical Formulation of the Problem

A general LPP can be stated as (Max/Min)Z = c1x1 + c2x2 + ........ + cnxn subject to givenconstraints and the non-negative restrictions.

x1, x2, ........, xn 0 and all are variables.

c1, c2, ........ cn are constants.

Graphical methods to solve a LinearProgramming Problem

Corner Point method: This method is used to solvethe LPP graphically by finding the corner points.

Procedure-

(i) Replace the signs of inequality by the equalityand consider each constraint as an equation.

(ii) Plotting each equation on the graph that willrepresent a straight line.

(iii)The common region that satisfies all theconstraints and the non-negative restrictionsis known as the feasible region. It is a convexpolygon.

(iv) Determining the vertices of the convex polygon.These vertices of the polygon are also knownas the extreme points or corners of the feasibleregion.

(v) Finding the values of Objective function at eachof the extreme points. Now, finding the pointat which the value of the objective function isoptimum as that is the optimal solution of thegiven LPP.

General features of a LPP

The feasible region is always a convex region.

The maximum (or minimum) solution of theobjective function occurs at the vertex (corner)of the feasible region.

If two corner points produce the same optimum(maximum or minimum) value of the objectivefunction, then every point on the line segmentjoining these points will also give the sameoptimum ( maximum or minimum) value.

CHAPTER 12 : Linear Programming 12.225


TOPIC 1

4 Marks Questions1. A small firm manufactures necklace and bracelets.

The total number of necklaces and bracelets that

i t can handle per day is at most 24. I t takes 1

hour to make a bracelet and half an hour make a

neck l ace. The maxi mum number of hour s

avai lable per day is 16. I f the profi t on a necklace

i s Rs100 and t hat on a br acel et i s Rs300,

formulate an L.P.P for finding how many of each

should be produced dai ly to maximize the profi t?

I t is being given that at least one of each must be

produced.

[DELH I 2017]

2. Solve the fol lowing L.P.P. graphical ly:

Minimise 5 10Z x y

Subject to 2 120x y

Constraints: 60, 2 0 and x,y 0x y x y

[DELH I 2017]

3. A cooperat ive society of farmers has 50 hectaresof land to grow two crops A and B The profi t fromcrops A and B per hectare are est imated asRs. 10,500 B and Rs. 9,000 respect ively. To controlweeds a l iquid herbicide has to be used for cropsA and B at the rate of 20 l i t res and 10 l i t res perhectare, respect ively. Fur ther not more that 800l i t res of herbicide should be used in order topr ot ect f i sh and wi ldl i fe i s mor e impor t antdrainage from this land. Keeping in mind thatthe protect ion of fish and other wi ldl i fe is moreimpor tant than ear ing profi t ,how much landshould be allocated to each crop so as to maximizethe total profi t ? From an LPP from the aboveand solve i t graphical ly.

[ALL I NDI A 2013]

6 Marks Questions4. A manufacturer produces two products A and B.

Both the products are processed on two differentmachines. The avai lable capacity of fi rst machineis 12 hours and that of second machine is 9 hoursper day. Each unit of product A requires 3 hourson both machines and each uni t of product Brequires 2 hours on first machine and 1 hour onsecond machine, Each unit of product A is sold atRs 7 profi t and that of B at a profi t of Rs 4. Findthe product ion level per day for maximum profi tgraphical ly.

[DELH I 2016]

Different Types of Linear ProgrammingProblems

Manufacturing problems

In order to make maximum profit, determinethe number of units of different products whichshould be produced and sold by a firm when eachproduct requires a fixed manpower, machinehours, warehouse space per unit of the outputetc., in order to make maximum profit.

Diet problems

Determining the minimum amount of differentnutrients which should be included in a diet soas to minimize the cost of the diet.

Transportation problems

To find the cheapest way of transporting aproduct from factories situated at differentlocations to different markets.

Allocation problems

These problems are concerned with the allocationof a particular land/area of a company or anyorganization by choosing a certain number ofemployees and a certain amount of area tocomplete the assignment within the requireddeadline, given that a single person works ononly one job within the assignment.

Integrat ion is the inverse of different iat ion. Insteadof diff

CHAPTER 13 : Probability13.236

[TOPIC 1] Conditional Probability andIndependent Events

Summary Probability:

Let S be the sample space and E be the event in anexperiment.

Then,

Number of favourable eventProbability

Total number of events = P E =

n E=

n S

Where, 0 n (E) n(s)

0 P (E) 1

Hence, the probability of the occurrence of an eventE is denoted by P(E)

Now, P E = 1- P E ( P E can also be written

as P(E'))

If probability of any event is one, this does notdepict the certainty of that event.

In similar way if the probability of any event iszero, this does not depict that the event willnever occur.

Mutually Exclusive Event: The two eventswhich cannot occur simultaneously are calledmutually exclusive events.

Let B = {1, 2, 3, 4, 5, 6}

X = the event of occurrence of a number greaterthan 5 = {6}

Y = the event of occurrence of an even number= {2, 4, 6}

Here, events X and Y are not mutually exclusivebecause they can occur together when thenumber 6 comes up.

Independent Events: If the occurrence or non-occurrence of one event is unaffected by theoccurrence or non-occurrence of other, theseevents are called independent events.

Consider an example of drawing two marblesone by one with replacement from ajar containing 2 red marbles and 1 yellowmarble

Now assume, X = the event of occurrence of ared marble in first draw

And Y = the event of occurrence of a yellowmarble in second draw

So, here the probability of occurrence Y is notaffected by that of X.

Hence, events X and Y are independent events.

Exhaustive Events: If the performance ofrandom experiment always results in theoccurrence of at least one of the given set ofevents, the set of those events will be known asexhaustive.

If their union is the total sample space

If event A, B and C are disjoint pairs i.e.,

Consider an example of throwing a die,A = {1, 2, 3, 4, 5, 6}

Now assume X = the event of occurrence of anmultiple of 2 = {2, 4, 6}

Y = the event of occurrence of the number notdivisible by 2 = {1, 3, 5}

Z = the event of occurrence of multiple of3 = {3,6}

Here X and Y are mutually exclusive but Y andZ are not.

Conditional Probability:

The probability of occurrence of event A when Bhas already been occurred is known as Conditionalprobability also called probability of occurrence ofA w.r.t B.

Some important formulae related to conditionalprobability

P A B

P A|B = , B i.e., P B 0P B

CHAPTER 13 : Probability 13.237

P A B

P B|A = , A i.e., P A 0P A

P A B

P A|B = , P B 0P B

P A B

P A|B = , P B 0P B

P A B

P A|B = , P B 0P B

P A|B + P A|B = 1

Some formulae

. .,

P and

P A B P A P B P A B i e

A or B P A P B P A B

P A B C P A P B P C P A B

P B C P C A P A B C

only ( )

( but not )

P A B P B P B A P

B A P B P A B

only

(A but not )

P A B P A P A B P

B P A P A B

( ) neither nor

1

P A B P B A P A B

P A B


TOPIC 1

2 Marks Questions1. A black and a red die are rol led together. Find

the condi t ional probabi l i ty of obtaining the sum8, given that the red die resulted in a numberless than 4.

[DELH I 2018]

2. Pr ove t hat i f E and F ar e i n dependen tevent s, t hen t he event s E and F ' ar e al soindependent .

[DELH I 2017]

3. A die, whose faces are marked 1, 2, 3 in red and4, 5, 6 in green, is tossed. Let A be the event"number obtained is even" and B be the event"number obtained is red". Find i f A and B areindependent events.

[ALL I NDI A 2017]

4 Marks Questions4. Find the mean number of heads in three tosses

of a fair coin

[ALL I NDI A 2011]

5. A and B throw a pair of dice al ternately, t i l l oneof them gets a total of 10 and wins the game.Find their respect ive probabi l i t ies of winning, i fA star ts fi rst .

[ALL I NDI A 2016]

6. Assume that each born chi ld is equal ly l ikely tobe a boy or a gi r l . I f a family has two chi ldren,what is the condit ional probabi l i ty that both aregir ls? Given that

(i ) the youngest is a gi r l

(i i ) at least one is a gir l .

[DELH I 2014]

7. Pr obabi l i t i es of sol vi ng a speci f i c pr obl em

independently by A and B are 1 12 3

and respectively

i f both t ry to solve the problem independent ly,find the probabi l i ty that (i) the problem is solved(2) exact ly one of them solves the problem.

[DELH I 2011]

CHAPTER 13 : Probability 13.241

[TOPIC 2] Baye's Theorem andProbability Distribution

Summary BAYES' theorem:

If E1, E2, E3 ……….. En are n non-emptyconstituting a partition of sample space S i.e.,S1, S2, S3 ……. Sn are pair wise disjoint and

1 2 3 nE E E ..... E = S and A is any event of

non- zero probability, then

1

E . |E| , 1,2,3,.......

E |E

i in

j jj

P P AP E A i n

P P A

For example,

1 11

1 1 2

2 3 3

E . |E| ,

E . |E E .

|E E . |E

P P AP E A

P P A P

P A P P A

1,2,3,.......i n

It is also known as the formula for the probabilityof cause.

Prior probabilities are the probabilities whichare known before the experiment takes place.

P(A|En) are called posterior probabilities.

Random Variable:

A real valued function defined over the samplespace of an experiment is known as randomvariable. It is denoted by uppercase lettersX, Y, Z etc.

Discrete random variable : When only finiteor countably infinite number of values can betaken by the random variable then it is calleddiscrete random variable.

Continuous random variable: When anyvalue between two given limits can be taken bythe variable then it is called continuous randomvariable.

If the values of a random variable together withthe corresponding probability are known, thenthis is called the probability distribution ofthe random variable.

Formulae:

Mean or Expectation of a random variable

X =

1n

i iiX x P

Variance =

2 2 2

1

n

i ii

P x

Standard deviation = Variance

Bernoulli Trials:

They are basically known as trials of a randomexperiment.

If they satisfy the following conditions:

There should be a finite number of trials.

The trials should be independent.

Each trial has exactly two outcomes: success orfailure.

The probability of success remains the same ineach trial

Binomial distribution:

A Binomial distribution with probability of successin each trial as p and with n Bernoulli trials isdenoted by B(n, p)

n and p are the parameters of Binomial Distribution

Therefore the expression P(x = r) or P(r) iscalled the probability function of BinomialDistribution.

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